Question

If x^(y)=5^(x-y) then dy/dx=​

Answer 1

First, rewrite

`x^y=e^(\ln x^y)=e^(y\ln x)`

`5^(x-y)=e^{\ln5^(x-y)} = e^(\ln(5)(x-y))`

Now, differentiate both sides using the chain rule:

`(\mathrm d\left(e^(y\ln x)\right))/(\mathrm dx)=(\mathrm d\left(e^(\ln(5)(x-y))\right))/(\mathrm dx)`

`e^(y\ln x)(\mathrm d(y\ln x))/(\mathrm dx)=e^(\ln(5)(x-y))(\mathrm d(\ln(5)(x-y)))/(\mathrm dx)`

`x^y\left((\mathrm dy)/(\mathrm dx)\ln x+y(\mathrm d(\ln x))/(\mathrm dx)\right)=\ln(5)5^(x-y)\left((\mathrm d(x))/(\mathrm dx)-(\mathrm dy)/(\mathrm dx)\right)`

`x^y\left(\ln x(\mathrm dy)/(\mathrm dx)+\frac yx\right)=\ln(5)5^(x-y)\left(1-(\mathrm dy)/(\mathrm dx)\right)`

`\left(x^y\ln x+\ln(5)5^(x-y)\right)(\mathrm dy)/(\mathrm dx)=\ln(5)5^(x-y)-yx^(y-1)`

`(\mathrm dy)/(\mathrm dx)=(\ln(5)5^(x-y)-yx^(y-1))/(x^y\ln x+\ln(5)5^(x-y))`